Question

A nutritionist wants to determine how much time nationally people spend eating and drinking. The record shows that the mean was 1.5 hours. Suppose for a random sample of 921 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.62 hours with a standard deviation of 0.71 hour. What is the upper boundary of the 95% C.I.?

Answer 1

Solution :

Given that,

= 1.62

s = 0.71

n = 921

Degrees of freedom = df = n - 1 = 921 - 1 = 920

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,920 =1.962

Margin of error = E = t_/2,df * (s /n)

= 1.962 * (0.71 / 921 )

= 0.046

Margin of error = 0.046

The 95% confidence interval estimate of the population mean is,

- E < < + E

1.62- 0.046 < < 1.62 + 0.046

1.574 < < 1.666

(1.574, 1.666 )